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Old Unix and Linux 03-19-2010
kelseyh kelseyh is offline
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Controlled For Loop

I have a for loop and i pass in number of times i want to loop as argument to script.


I have attached a snippet of my script. Problem is this does not work, if I manually replace the $NO_OF_LOOPS with a number in the script it works fine.

NO_OF_LOOPS=$1
for i in {0..$NO_OF_LOOPS}

Any help, i just want to control the no of loops via the argument i pass in.
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Old Unix and Linux 03-19-2010
thillai_selvan's Unix or Linux Image
thillai_selvan thillai_selvan is offline
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How about this!

Code:
NO_OF_LOOPS=$1
for i in `seq 0 $NO_OF_LOOPS`
do
        echo $i
done

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Old Unix and Linux 03-19-2010
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Code:
 
NO_OF_LOOPS=$1
for i in `seq  0 $NO_OF_LOOPS`

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Old Unix and Linux 03-19-2010
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Quote:
Originally Posted by murugaperumal View Post
Code:
 
NO_OF_LOOPS=$1
for i in `seq  1 NO_OF_LOOPS`

This will not work.
Error:
seq: invalid floating point argument: _OF_LOOPS
Try `seq --help' for more information.

You need to precede $ infront of NO_OF_LOOPS

It should be like

Code:
for i in `seq  1 $NO_OF_LOOPS`

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Old Unix and Linux 03-19-2010
murugaperumal's Unix or Linux Image
murugaperumal murugaperumal is offline
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Quote:
Originally Posted by thillai_selvan View Post
This will not work.
Error:
seq: invalid floating point argument: _OF_LOOPS
Try `seq --help' for more information.

You need to precede $ infront of NO_OF_LOOPS

It should be like

Code:
for i in `seq  1 $NO_OF_LOOPS`


Now I corrected
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Old Unix and Linux 03-19-2010
amitranjansahu's Unix or Linux Image
amitranjansahu amitranjansahu is offline
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Or in bash


Code:
for ((  i = 1 ;  i <= $1;  i++  ))
do
	echo "amit $i"
done

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Old Unix and Linux 03-19-2010
kelseyh kelseyh is offline
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I get seq: command not found
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